Calculating an Escape from Earth's Gravity: Can It Be Done?

[Salvador]

this is an interesting thought i had but i don't have the mathskills to calculate the answer so I thought I could ask for your help.

So I want to drive my car up to space for example , imagine I have built a ramp on which the acceleration goes in the opposite direction of the rotation of earth, (to gain more tangential velocity ) Now here are the variables, the ramp reaches certain heigth X the car is traveling at certain speed up the ramp , how high the ramp and how fast the speed of the car or object traveling would need to be in order to achieve a speed which would be high enough to break away from Earth's gravity and adding the speed + the tangential velocity of Earth so that the car could fly off into space without being dragged back down? I guess the angle of the ramp would also have to be considered.

This is surely not a homework question even though at some points it sounds similar to one.

thanks for interest.

 

[axmls]

Well, to maintain a low-earth orbit, you need an orbital velocity of 7.8 kilometers per second, for starters.

 

[CWatters]

Lets ignore the car. It's not really fast enough to make much of a difference.

If you just built a rigid ramp or tower the Earth's rotation would make the top move faster than the bottom. The higher it is the faster the top would move.

Unfortunately at low Earth orbit the top of the tower wouldn't be fast enough for you to step off and stay in orbit. The difference in velocity is large so a car won't help you. However the required velocity goes down with height so if you keep building it taller and taller you would eventually reach a height at which you could just drive or walk up, step off and float in orbit.

The velocity of the top of the tower (in m/s) would be given by..

V = 2 π h/T

where
h = height of tower (in meters from centre of the earth)
T = Length of a day in seconds

The required orbital velocity is given by

V = Sqrt (GMe/h)

where
G = Gravitational constant
Me = Mass of the earth

If you equate the two you can solve for h.

I made h about 42m meters or 42,000 km from the centre of the earth.

The radius of the Earth (Re) is about 6400 km so the tower would need to be about 35,600 km tall.

Google suggests 35,786 km for the height of a space elevator so not too far out.

 

[CWatters]

PS The fastest rail guided passenger vehicle (aka the Maglev Train) is about 400 km/h. It would take about four days to climb up.

 

[Salvador]

the car was basically just a thing which I used to describe whatever I was thinking of shooting into space , not that a typical car can achieve anything close to those speeds.

well the maglev is fast but for such applications as we discuss here haven't some military " guys" already tried something like shooting payloads with rail guns or something?
As much as I remember using a linear motor especially with some specific designs one can achieve pretty outstanding speeds or accelerations those would definitely make the tomwer's height lower.?